Acar drives with a velocity of 35 miles/hour in a direction 37° south of east for 1.5 hours, then with a velocity of 24 miles/hour in a direction 58° south of west for 45 minutes, then with a velocity of 45 miles/hour due east for 1.2 hours. a) what is the magnitude and direction of the car's total displacement? b) in order to return to its point of origin, what is the magnitude and direction of the displacement that the car must follow?

a) The magnitude of the car´s total displacement is 99 mi and the direction is 29° south of east.

b) The magnitude of the return displacement is 99 mi and the direction is 29° north of west.

Explanation:

To find the magnitude of the total displacement vector (vector R in the figure) we have to sum the displacement vectors a, b, and c and then calculate the magnitude of the resulting vector.

First, let´s find the magnitude of each displacement vector using the following equation:

x = v · t

Where:

x =  displacement

v = velocity

t = time

Then, the displacements will be:

Magnitude of a = 35 mi/h · 1.5 h = 53 mi

Magnitude of b = 24 mi/h · 0.75 h = 18 mi

Magnitude of c = 45 mi/h · 1.2 h =54 mi

Now, using trigonometry we can find the components of each vector.

Vector a:

If ax is the x-component of the vector a, by trigonometry, we know that

cos 37° = magnitude of ax / magnitude of a

magnitude of a · cos 37° = magnitude of ax

ax = 53 m ·  cos 37° = 42 mi  

For the y-component:

ay = 53 m · sin 37° = 32 mi    

Then, considering east and north as positive directions the vector a will be:

a = (42 mi, -32 mi)

In the same way for the other vectors

Vector b:

bx = b · cos 58°

bx = 18 mi · cos 58° = 9.5 mi

by= 18 mi · sin 58° = 15 mi

Then,

b =(-9.5 mi, -15 mi)

Vector c

cx = 54 mi · cos 0° = 54 mi

cy = 54 mi · sin 0° = 0

c = (54 mi, 0)

The sum of the displacement vectors will give the car´s total displacement:

R = a + b + c

R = (42 mi, -32 mi) + (-9.5 mi, -15 mi) + (54 mi, 0)

R =(42 mi - 9.5 mi + 54 mi, -32 mi - 15 mi + 0) = (87 mi, -47 mi)

The magnitude of R is calculated as follows:

R² = (87 mi)² + (-47 mi)²

R = 99 mi

The direction of the vector R is calculated using trigonometry:

cos α = magnitude of x-component of R / magnitude of R

(Where α is the angle measured south of east)

Then:

cos α = 87 mi/99 mi = 29°

Then, the direction of the total displacement vector is 29° south of east.

b) To return to the point of origin, we have to find the vector R1 that added to the total displacement vector R will give the null vector (0 , 0). Then:

R + R1 = (87 mi, -47 mi) + (r1x + r1y) = (0, 0)

R + R1 =(87 mi + r1x, -47 mi + r1y) = (0, 0)

Then:

r1x = -87 mi

r1y = 47 mi

R1 = (-87 mi, 47 mi)  (see R1 in the figure)

The magnitude will be the same as the magnitude of R, 99 m.

The direction is calculated in the same way as it was calculated for R only that in this case, we will obtain the north of west angle:

cos α = 87 mi/99 mi

α = 29°

Then, the magnitude of the displacement that the car must follow to return is 99 mi and the direction is 29° north of west.

probably not if net force is zero

massxgravity x heght=weightxheight= pe

4900=490xh

h=10